Explore topic-wise InterviewSolutions in .

Given that,

D = 14 CM, So, R = 14/2 = 7 cm

We know that,

Let the length of the required SIDE be x cm

Area of the trapezium = [$(\FRAC{1}{2})$ × (sum of parallel sides) × (Distance between parallel sides)]

One of the parallel side = 0.1 m = 10 cm

Then, area of the trapezium = {¹/? × (10 + x) × 4} cm²

= (20 + 2x) cm²

But, the area of the trapezium = 30 cm² (given) 

⇒ 20 + 2x = 30 

⇒ 2x = 30 - 20 

⇒ 2x = 10 

⇒ x = 10/2 

⇒ x = 5

From the given data,

No of sides n = 20

Perimeter of the POLYGON i.e, sum of the total sides s = 2600

Let us consider smallest side be a

Largest side is 12 times that of the smallest side

∴ largest side = 12 a

By using sum of A.P formula, we get

Largest side = a + (n - 1) d = 12 a

Sum = n/2 × [2a + (n - 1)d] = n/2 × [ a + a + (n - 1)d] = n/2 × [a + 12a]

s = n/2 × (13a)

By substituting the value of s = 2600 and n = 20 ,we get

⇒ 2600 = 20/2 × (13a)

⇒ 2600 = 10 (13a)

⇒ 260 = 13 a

⇒ smallest side a = 20 cm and largest side 12a = 12 × 20 = 240 cm

We KNOW that largest side = a + (n - 1)d = 12a = 240

⇒ 240 = 20 + (20 - 1)d

⇒ 220 = 19 d

⇒ d = 220/19 = 11.57 cm

Area of the square = 484 SQ. cm

Let each SIDE B x cm

X² = 484

⇒ x = √484 = 22 cm

∴ Length of the wire = 4 × 22 = 88 cm

∴ Circumference of the desired circle = 88 cm

⇒ 2 × 22/7 × radius = 88

⇒ radius = 14 cm

LET SIDE of equilateral triangle is a

Area of equilateral triangle = 25√3

⇒ √3/4 × a² = 25√3

⇒ a² = 100

CURVED SURFACE AREA of a HEMISPHERE = 2πr²

Volume of a hemisphere = (2/3)πr³

110.88/155.232 = 2πr²/(2/3)πr³

r = 4.2 CM

First we need to calculate the area of RHOMBUS

Area of rhombus = 1/2 × d₁ × d₂ = 1/2 × 36 × 24 = 432 m²

There are 1200 TIMES which are required for decoration purpose

Area of 1200 tiles = 1200 × 432

⇒ Area of 1200 times = 518400 m²

The diameter of a SPHERE is equal to the diameter of another sphere.

Let the radius of FIRST sphere be x cm and radius of second sphere will ALSO be x cm.

The curved surface area of first sphere = 4πr²[Where, r = radius]

= 4π × (x)² = 4πx² cm²

The volume of the second sphere = (4/3) × π × r³ [Where, r = radius]

= (4/3) × π × x³ cm³

The curved surface area of the first and the volume of the second are NUMERICALLY equal.

4πx² = (4/3) × π × x³

⇒ x = 4 × (3/4)

⇒ x = 3

Volume of hemisphere = (2/3) × πr³

Where, r = radius of hemisphere = 21 cm

⇒ Volume of hemisphere = (2/3) × (22/7) × 21 × 21 × 21 = 19404 cm³

After conversing into LITERS,

⇒ 19404 cm³ = 19.404 liters

For TOTAL volume of mixture,

⇒ Total volume of mixture = 19.404 × 200 = 3880.8 liters

For REQUIRED solution,

⇒ Required mixture = 3880.8 – 10 = 3870.8 liters

⇒ Height of the trapezium (h) = 20 cm

⇒ Area of the trapezium (A) = 330 cm²

⇒ Ratio of the parallel side = 4 : 7 = 4x & 7x

⇒ Area of trapezium (A) = 1/2 (sum of parallel side) × height

⇒ 330 = 1/2(4x + 7x) × 20

⇒ 330 × 2 = 11x × 20

⇒ 660/20 = 11x

⇒ 33/11 = x

⇒ x = 3

⇒ Shorter side = 4x = 4 × 3 = 12 cm

⇒ LONGER side = 7x = 7 × 3 = 21 cm

⇒ Difference = 21 – 12 = 9 cm

A rhombus has 2 congruent opposite acute angles and TWO congruent opposite OBTUSE angles. One of the properties of a rhombus is that any two internal consecutive angles are supplementary. Let x be the acute angle. The obtuse angle is twice: 2x. Which gives the FOLLOWING equation.

x + 2 x = 180 degrees.

Solve the above equation for x.

3x = 180 degrees.

x = 60 degrees.

We use formula 2 for the area of a rhombus knowing the side length and one of the angles.

area of rhombus = (10 feet) 2 SIN (60 degrees)

= 86.6 feet 2 (rounded to 1 decimal PLACE)

[Updated by Reviwer]

AREA of circle of RADIUS 14 cm = πr² = 22/7 × 14 × 14 = 616 cm²

Area of semi-circle of radius 3.5 cm = πr²/2 = 22/7 × 3.5 × 3.5/2 = 19.25 cm²

∴ MAXIMUM no. of semi-circles that can be cut = 616/19.25 = 3

Let a be the apothem LENGTH, base length, SLANT height and height are b, s and h respectively.

Surface area of a hexagonal pyramid = 3ab + 3bs

⇒ 3 × 5 × 6 + 3 × 6 × 12

⇒ 90 + 216 = 306cm²

Volume of the pyramid = abh

⇒ 5 × 6 × 11 = 330cm³

A regular TETRAHEDRON is a triangular PYRAMID in which all the four FACES are equilateral triangles

Given, edge length = a = 6√2 cm

? Volume of regular tetrahedron = (edge)3$/6√2

Let side of the cube is a.

Total surface AREA of the cube = 6a²

⇒ 6a² = 2904

⇒ a² = 2904/6 = 484

⇒ a = 22 cm

Solution :

Given:

Perimeter of five squares $P1 = 24cm , P2 = 32 cm, P3 = 40 cm , P4 = 80 cm, P5 = 100 cm$.

Perimeter of a square $= 4a$ units ,where $'a'$ is the side of the square. 

So, Side of the five squares are,

$a1 = 24/4 = 6$ cm 

$a2 = 32/4 = 8$ cm

$a3 = 40/4 = 10 $cm

$a4 = 80/4 = 20 $cm

$A5 = 100/4= 25$ cm

Area of a square $=a^2$ sq.units 

So,the area of five squares are

$A1 = 6^2 = 36$ sq.cm 

$A2 = 8^2 = 64$ sq.cm

$A3 = 10^2 = 100$ sq.cm

$A4 = 20^2 = 400$ sq.cm

$A5 = 25^2 = 625$ sq.cm

It is given that,

Area of the 6TH square = Sum of the area of five squares

$A6 = 36 + 64 + 100 + 400 + 625$

$A6 = 1225$ sq.cm.

$=> a6^2 = 1225$

$a6 = \sqrt{1225}$

$a6 = 35$

So, Perimeter of the $6th$ square $= 4a$ units

$P6 = 4 × 35$ cm

$P6 = 140$ cm

So, the perimeter of 6th square is $140$ cm .

The correct option is 3).140 cm

Let the side of the ORIGINAL cube = ‘X’ cm

Side of the new cube = (x - 4) cm

Difference between the original and new volume = 604

x³ - (x - 4)³ = 604

⇒ x³ - (x³ + 48X - 12x² - 4³) = 604

⇒ x³ - (x³ + 48x - 12x² - 64) = 604

⇒ x³ - x³- 48x + 12x² + 64 = 604

⇒ 12x² - 48x + 64 - 604 = 0

⇒ 12x² - 48x - 540 = 0

⇒ x² - 4x - 45 = 0

⇒ x² - 9X + 5x - 45 = 0

⇒ x(x - 9) + 5(x - 9)= 0

⇒ (x + 5) (x - 9) = 0

⇒ (x + 5) = 0 or (x - 9) = 0

⇒ x = -5 or x = 9

Volume = (Side)³ = (9)³ = 729 cm³

∴ Volume of the original cube = 729 cm³

Let the time taken to fill the cistern be‘t’ hr.

Given, Diameter of cistern is 10m and depth is 2m.

We know that, Volume of CYLINDER = πR² × h

⇒ Radius of cylinder = Diameter/2

= 5 m

∴ Volume of cylinder = π × 5² × 2 m³

= 50π m³

Given, Velocity of WATER = 3 km/hr

= 3000 m/hr (? 1 km = 1000 m)

Inner diameter of pipe = 20 cm

∴ Inner radius = 20/2 = 10 cm

= 0.1 m (? 1m = 100 cm)

∴ Volume flow rate of water = π × (inner radius)² × velocity of water

= π × 0.1² × 3000 m³/hr

= 30π m³/hr

Now,

⇒Time taken to fill the cistern $(,t = \frac{{Volume\ of\ cistern}}{{Volume\ flow rate}})$

$(\Rightarrow t = \frac{{50{\rm{\pi }}}}{{30{\rm{\pi }}}}\ {\rm{hr}})$

⇒ t = 1.67 hours

Area of circle with RADIUS ‘r’ = πr2$

⇒ 22/7 × r2$ = 3850

⇒ r2$ = 1225

⇒ r = √1225 = 35 cm

1 : 7 : 19

Since we KNOW that ratio of perimeters of two TRIANGLES is same as that of sides

⇒ Perimeter of ?XYZ/Perimeter of ?PQR = 16/9 = XY/PQ

⇒ XY = 16 × 3.6/9 = 6.4 CM

∴ XY is 6.4 cm

We KNOW that,

order of rotational symmetry is the property a shape has when it looks the same after some rotation by a PARTIAL TURN.

? Trapezium has only one pair of parallel sides, it has no LINES of symmetry

∴ The order of rotational symmetry of a trapezium is 1

We know, total surface AREA of a HEMISPHERE = 3πr²

So, total surface area of a hemisphere = 1039.5

⇒ 3πr² = 1039.5

⇒ 3 × 22/7 × r² = 1039.5

⇒ r² = 110.25

⇒ r = 10.5

Let the length and breadth of the field be ‘l’ m and ‘B’ m respectively.

According to the question, the length of the field is 3 m more than twice the breadth

⇒ l = (2b + 3) m

We KNOW that,

PERIMETER of a rectangular field = 2(l + b)

⇒ 84 = 2(2b + 3 + b)

⇒ 84/2 = 3b + 3

⇒ b = 39/3 = 13 m

∴ b = 13 m

l = 2b + 3 = 2 × 13 + 3 = 29 m

∴ l = 29 m

In a kite Two pairs of CONSECUTIVE sides are congruent, ONE - PAIR of diagonally opposite ANGLES is equal, only one DIAGONAL is bisected by the other and the diagonals cross at 90°.

Area of EQUILATERAL triangle of side 10 CM can be GIVEN as,

⇒ Area of triangle = √3 × (10)²/4 = 25√3

∴ the area of the triangle is 25√3 cm²

SIDE can be GIVEN as Diagonal/√2

⇒ Side = 12/√2 = 6√2

∴ The side of SQUARE is 6√2

Radius of cylinder = r = 21/2 = 10.5 cm

Height of cylinder = H = 10 cm

Now, VOLUME of cylinder = πr²h

AREA of a trapezium = ½ × HEIGHT of trapezium × (sum of LENGTHS of parallel sides)

⇒ 165 = ½ × 5.5 × (sum of lengths of parallel sides)

⇒ Sum of lengths of parallel sides = (165 × 2)/5.5 = 60 cm

We know that area of trapezium = 1/2 × HEIGHT × (a + b)

Where, a = LENGTH of the side parallel to the base

b = base

⇒ 18 sq.cm = 1/2 × 3 × (a + 5)

⇒ 36 = 3a + 15

⇒ 21 = 3a

⇒ a = 21/3 = 7 cm

Volume of cylinder = πr²h = π × 8 × 8 × 72 = (72 × 64) π cm³

Volume of SPHERE = (4/3) πr³ = (4/3) π × 4 × 4 × 4 = (256/3) π cm³

Number of spherical bearings = Volume of cylinder/ volume of sphere = (72 × 64) π cm³/(256/3) π cm³

Height will be divided in 3 parts and heights for three cones will be 7cm, 14cm and 21cm, in the ratio (1 ? 2 ? 3)

Also the RADIUS of the three cones will be 3CM, 6cm and 9cm, in the ratio (1 ? 2 ? 3)

Since curved surface area of a cone = πrl

⇒ Ratio of curved surface AREAS of three cones = r_1l$1 ? r_{2l$2 ?} r_3l$3 = 21 : 90 : 198

The volumes of the SPHERE and of the PIPE will be the same.

Diameter of sphere = 6 cm

∴ RADIUS of sphere = 6/2 = 3 cm

Volume of the sphere = 4πr³/3 = 4π(3)³/3 = 36π

Let the internal radius of pipe be r.

Volume of the hollow cylindrical pipe = πh (R² – r²) where h= HEIGHT, R = external radius= 10/2 = 5 cm.

Hence,

36 π = π (4)(5² – r²)

⇒ 25 – r² = 9

⇒r² = 16

⇒ r = 4 cm

A SPHERICAL lead ball of radius 6 cm is melted and small lead balls of radius 3 mm are made.

We know,

VOLUME of sphere = (4/3) × π × r³ [Where r = radius]

6 cm = 60 mm

So, the volume of spherical lead ball of radius 60 mm = (4/3) × π × 60³ = 288000π mm³

And the volume of the small lead balls of radius 3 mm = (4/3) × π × 3³ = 36π mm³

Hence, the total NUMBER of possible small lead balls = 288000π/36π = 8000.

Let the radius of Sphere be r cm

Then CSA of sphere is 4πr²

If R is the radius of cone and ‘L’ is the slant HEIGHT then CSA of cone = πRl

l = √(R² + h²) = √(30² + 40²) = 50 cm

Given CSA of Sphere = CSA of cone

⇒ 4 × π × r² = π × 30 × 50

⇒ r² = 15 × 25 = 375

Side of the square BASE = 8 cm and HEIGHT of the pyramid = 16 cm

SLANT height (L) of the pyramid needs to be evaluated,

⇒ l² = (side/2)² + (Height)²

⇒ l² = 16 + 256 = 272

⇒ l = 4√17 cm

∴ Total surface area of the pyramid = (Base Area) + ½ × (Perimeter of base) × (Slant height)

⇒ 64 + ½ × 32 × 4√17

⇒ 64 + 64√17

Let the radius of the upper base of the frustum be ‘r’ cm

Now,

⇒ Volume of CONE = (1/3) × π × (base radius)² × height

⇒ Volume of cone before cutting = (1/3) × π × (15)² × 40 = 3000π cm³

Now, when a cone is cut parallel to its base, the base part is the frustum and the remaining part is ALSO a cone, whose base radius is equal to the radius of the upper base of the frustum

⇒ Base radius of remaining cone = r cm

As,

⇒ Volume of frustum = 2000π cm³

⇒ Volume of other cone = 3000π – 2000π = 1000π cm³

Also,

⇒ Height of other cone = 40 – 10 = 30 cm

According to the question,

⇒ 1000π = (1/3) × π × (r)² × 30

⇒ r² = 100

⇒ r = √100 = 10 cm

Let a, b and c be the LENGTHS of the triangle.

Let ‘a’ be the greatest side with LENGTH 17 cm and c be the smallest length.

Given, Perimeter of triangle = 40 cm

⇒ (sum of all sides) = 40

⇒ a + b + c = 40

⇒ 17 + b + c = 40

⇒ b + c = 23

Then semi-perimeter:

⇒ s = Perimeter/2

= 40/2 cm

= 20 cm

We know that,

Area of triangle = $(\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)})$

Where, s is semi perimeter and a, b and c are sides of the triangle

$(\Rightarrow 60 = \sqrt {20 \times \left( {20-17} \right) \times \left( {20-b} \right) \times \left( {20-\left( {23-b} \right)} \right)})$

Squaring both sides

⇒ 3600 = 60 × (20 - b) × (b - 3)

⇒ 60 = -b² + 23b – 60

⇒ b² – 23b + 120 = 0

Solving above we get,

b = 15 or 8

So if b = 15CM then,

⇒ c = 23 – 15

= 8 cm

And if b = 8 cm then,

⇒ c = 23 – 8

= 15 cm

As we know,

SURFACE area of SPHERE = 4π r², where r is the radius of sphere

Surface area of cube = 6a², where a is side of cube

∴ Surface area of sphere = surface area of cube

⇒ 4πr² = 6a²

$(\begin{array}{l} \Rightarrow \;\frac{{{r^2}}}{{{a^2}}} = \frac{6}{{4\pi }}\\ \Rightarrow \frac{r}{a} = \SQRT {\frac{3}{{2\pi }}}\end{array})$

Now,

As we know,

VOLUME of sphere = 4π r³/3, where r is the radius of sphere

Volume of cube = a³, where a is side of cube

∴ (volume of sphere)/(Volume of cube) $(= \;\frac{{\frac{4}{3}\pi {r^3}}}{{{a^3}}})$

$(= \frac{{4\pi }}{3} \times {\left( {\frac{r}{a}} \right)^3} = \;\frac{{4\pi }}{3} \times {\left( {\sqrt {\frac{3}{{2\pi }}} } \right)^3} = \frac{{4\pi }}{3} \times \frac{3}{{2\pi }} \times \sqrt {\frac{3}{{2\pi }}}= \;\sqrt {\frac{6}{\pi }} )$

AREA of trapezium = (½)(Sum of parallel sides) × (Distance between parallel sides)

⇒ Let the side of an equilateral TRIANGLE be a

⇒ Area of equilateral triangle = (√3 a²)/4

⇒ 121√3 = (√3 a²)/4

⇒ a² = 121 × 4

⇒ a = 22

⇒ Total length of wire = 3A = 3 × 22

⇒ Total length of wire = 66

⇒ when it bent into CIRCLE then circumference = 66

⇒ 2πr = 66

⇒ r = (33/22) × 7

⇒ r = 10.5 cm

⇒ Area = πr²

⇒ Area = (22/7) × 10.5²

Let n such CUBES be used.

Thus total VOLUME of all the cuboids:

⇒ n × 10 × 20 × 30

⇒ 6000n

This is also the volume of the final CUBE and that is of the form a³.

∴ n = 36.

Let the length and BREADTH of the rectangular field are ‘l’ and ‘b’ RESPECTIVELY. Then,

According to the question,

Length of the field is increased by 15%-

⇒ New length = 115% of l = (23/20)l,

Breadth of the field is increased by 10%-
⇒ New breadth = 110% of b = (11/10)b

We KNOW that,

Area of a rectangle = Length × Breadth

∴ New area of the rectangle = (23/20)l × (11/10)b

= (253/200) × l × b

Percent increase in the area-

$(= \frac{{\frac{{253}}{{200}}lb - lb}}{{lb}} \times 100\%= \left( {\frac{{253}}{{200}} - 1} \right) \times 100\%= \frac{{53}}{{200}} \times 100\%= 26.5\%)$

A CYLINDER of radius 4.3 cm and HEIGHT 4.8 cm is melted and constructed into a cone of the same radius.

So, the volume of the cylinder = π × r² × h = (22/7) × 4.3² × 4.8 cm³

Let the height of the cone be h.

So, the volume of the cone = (1/3) × π × r² × h = (1/3) × (22/7) × 4.3² × h cm³

Now we can WRITE,

(1/3) × (22/7) × 4.3² × h = (22/7) × 4.3² × 4.8

⇒ h = 4.8 × 3

⇒ h = 14.4

Solution;

GIVEN:

HEIGHT of the cylinder, $H = 8$ cm

Base radius of the cylinder, $r = 6$ cm

Radius of the CONE,$r = 6$ cm 

Height of the cone, $h = 8$ cm

Therefore, slant height $l^2 = h^2 + r^2$

$l^2 = 8^2 + 6^2$

$l^2 = 64 + 36$

$l^2 = 100$

$l = (100)^{1/2}$

Slant height, $l = 10$ cm

TSA of remaining solid

= CSA of cylinder + Area ofbase of cylinder + CSA of cone

 $= 2 \pi r h + \pi r^2 + \pi r l$

$= \pi r [ 2h + r + l ]$

$= \pi 6 [ 2×8 + 6 + 10 ]$

$ =\pi 6 [ 16 + 16 ]$

$= \pi( 6 × 32)$

$= 192 \pi$ sq.cm.